Exercise 24 Page 172

The given system consists of equations of planes. When using the Substitution Method to solve a system of equations, it is necessary to isolate a variable. In the third equation,

z

is already isolated, so we will start by substituting its value into the second equation.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 13 = 3 x - y 4 y - 3 x + 2 z = - 3 z = 2 x - 4 y

Substitute

$(II):$ $z = 2 x - 4 y$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 13 = 3 x - y 4 y - 3 x + 2 (2 x - 4 y) = - 3 z = 2 x - 4 y

▼

(II):

Solve for

x

Distr

$(II):$ Distribute $2$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 13 = 3 x - y 4 y - 3 x + 4 x - 8 y = - 3 z = 2 x - 4 y

AddSubTerms

$(II):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 13 = 3 x - y x - 4 y = - 3 z = 2 x - 4 y

AddEqn

$(II):$ $LHS + 4 y = RHS + 4 y$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 13 = 3 x - y x = 4 y - 3 z = 2 x - 4 y

This time, the

x -

variable was isolated in the second equation. We can now substitute its equivalent expression into the first equation.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 13 = 3 x - y x = 4 y - 3 z = 2 x - 4 y

Substitute

$(I):$ $x = 4 y - 3$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 13 = 3 (4 y - 3) - y x = 4 y - 3 z = 2 x - 4 y

▼

(I):

Solve for

y

Distr

$(I):$ Distribute $3$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 13 = 12 y - 9 - y x = 4 y - 3 z = 2 x - 4 y

SubTerm

$(I):$ Subtract term

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 13 = 11 y - 9 x = 4 y - 3 z = 2 x - 4 y

AddEqn

$(I):$ $LHS + 9 = RHS + 9$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 22 = 11 y x = 4 y - 3 z = 2 x - 4 y

DivEqn

$(I):$ $LHS / 11 = RHS / 11$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 = y x = 4 y - 3 z = 2 x - 4 y

RearrangeEqn

$(I):$ Rearrange equation

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = 2 x = 4 y - 3 z = 2 x - 4 y

The value of

y

is

2 .

Substituting

2

for

y

into the second equation, we can find the value of

x .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = 2 x = 4 y - 3 z = 2 x - 4 y

Substitute

$(II):$ $y = 2$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = 2 x = 4 (2) - 3 z = 2 x - 4 y

Multiply

$(II):$ Multiply

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = 2 x = 8 - 3 z = 2 x - 4 y

SubTerm

$(II):$ Subtract term

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = 2 x = 5 z = 2 x - 4 y

Now that we know the values of

x

and

y,

we are able to find the value of

z .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = 2 x = 5 z = 2 x - 4 y

SubstituteII

$(III):$ $x = 5$ , $y = 2$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = 2 x = 5 z = 2 (5) - 4 (2)

▼

(III):

Evaluate right-hand side

Multiply

$(III):$ Multiply

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = 2 x = 5 z = 10 - 8

SubTerm

$(III):$ Subtract term

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = 2 x = 5 z = 2

The solution to the system is the point

(5, 2, 2) .

This is the singular point at which all three planes intersect. Let's check our solution by substituting the values into the system.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 13 = 3 x - y 4 y - 3 x + 2 z = - 3 z = 2 x - 4 y

$(I), (II), (III):$ Substitute values

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 13 = ? 3 (5) - 2 4 (2) - 3 (5) + 2 (2) = ? - 3 2 = ? 2 (5) - 4 (2)

$(I), (II), (III):$ Multiply

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ 13 = ? 15 - 2 8 - 15 + 4 = ? - 3 2 = ? 10 - 8

$(I), (II), (III):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 13 = 13 ✓ - 3 = - 3 ✓ 2 = 2 ✓

Since the substitution of our answers into the given equations resulted in three identities, we know that our solution is correct.